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\title{Harvard Math 129: Algebraic Number Theory\\
\vspace{2ex}\\
{\Large\sc Sketches of Solutions to the Midterm}}
\author{William Stein}
\newcommand{\soln}[1]{{\par\sl #1}}
\begin{document}
\maketitle
These are sketches of solutions to the exam. I have written them so that
you, having worked on the exam, can easily see what the answer is to each
question. (Note: These solutions are slightly more abbreviated than I would have
liked for your solutions.)
\begin{enumerate}
\item (20 points) Let $R$ be a noetherian integral domain, and let $K=\Frac(R)$
be the field of fractions of $R$. Let $\Kbar$ be an algebraic
closure of $K$. Let $\overline{R}$ be the set of $\alpha \in \Kbar$
such that there is a nonzero monic polynomial $f(x) \in R[x]$ with
$f(\alpha)=0$. Is $\overline{R}$ necessarily a ring? Prove or
give a counterexample.
\soln{
This is true. The proof is exactly like the one we gave
in the book for $R=\Z$, but everywhere that we put~$\Z$
and $\Zbar$ there put $R$ and $\overline{R}$.
}
\item (10 points) Let $\O_K$ be the ring of integers of a number field~$K$,
and suppose $K$ has exactly $2s$ complex embeddings. Prove
that the sign of the discriminant of $\O_K$ is $(-1)^s$.
\soln{We proved in class that
$$
d_K = \det(A)^2 = ((-2i)^{-s} \cdot \Vol(V/\sigma(\O_K)))^2
$$
The right hand side is
$$
((-2)^{-s}\cdot \Vol(V/\sigma(\O_K)))^2 \cdot i^{-2s}
= \alpha \cdot i^{-2s} = \alpha \cdot (-1)^s,
$$
where $\alpha$ is a positive real number, since it is the
square of a real number. Thus the sign of $d_K$ is $(-1)^s$.
}
\item Suppose $K$ is a number field. For any
finite extension~$L$ of $K$, define set-theoretic maps
\begin{align*}
\Psi_L:&\,\, C_K \to C_L, \qquad [I] \mapsto [I \O_L]\\
\Phi_L:&\,\, C_L \to C_K, \qquad [I] \mapsto [I \cap \O_K],
\end{align*}
where $[I]$ denotes the class of the nonzero integral ideal~$I$.
\begin{enumerate}
\item (10 points) Is $\Psi_L$ a group homomorphism? Prove
or give a counterexample.
\soln{
Yes, $\Psi_L$ is a group homomorphism. It is well defined because
$I=(\alpha)$ maps to $\alpha\O_L$, which is principal.
It is a homomorphism since $I\O_L J\O_L = IJ\O_L$.
}
\item (10 points) Is $\Phi_L$ a group homomorphism? Prove
or give a counterexample.
\soln{
The map $\Phi_L$ is not even well defined, so it can't
be a group homomorphism. It does induce a
set-theoretic map on ideals. However, it need not send principal
ideals to principal ideals. For example, suppose $\wp\subset \O_K$
is a prime ideal that is not principal. Then (as we'll see below),
there is an extension $L$ of $K$ such that $I = \wp \O_L = (\alpha)$
is principal. But $I \cap \O_K = \wp$, since $\wp$ is contained
in this intersection, and $\wp$ is maximal. Thus the image of a principal
need not be principal.
}
\item (20 points) Prove that there is a number field $L$ such that
$\Psi_L$ is the~$0$ map, i.e., $\Psi_L$ sends every element of~$C_K$
to the identity of~$C_L$.
[Hint: Use finiteness of $C_K$ in two ways.]
\soln{
Let $I\in \O_K$ be a nonzero ideal.
The class group $C_K$ is finite, so there is an integer~$n$ such
that $[I]^n$ is trivial, so $I^n = (\alpha)$ is principal.
Let $L=K(\alpha^{1/n})$. Let $J= I \O_L$. Then
$$(J/(\alpha^{1/n}\O_L))^n = J^n / (\alpha\O_L) = (I^n\O_L) / (\alpha \O_L)
= \alpha \O_L / \alpha\O_L = \O_L,
$$
where we've used that our observation that the process of
extending an ideal to $\O_L$ preserves multiplication to
see that $J^n = I^n \O_L$. By unique factorization of ideals in
$\O_L$ and the fact that the fractional ideals are a torsion free
group, we see that $J/(\alpha^{1/n}\O_L)$ is the unit ideal (since
its $n$th power is the unit ideal). Thus $J=(\alpha^{1/n}\O_L)$
is principal, which proves the claim.
}
\end{enumerate}
{\small (Note: I wonder, is there always an $L$ such that $\Phi_L$ is the
$0$ map? This question just occured to me while writing
this exam. If you find an answer and tell me the answer,
I'll be very thankful, though this is not part of the
official exam. This question is related to ``visibility of
Mordell-Weil groups'', which I just wrote a paper about.)}
\soln{This question is meaningless since $\Phi_L$ is not well
defined. Also as mentioned above the induced map on sets of
fractional ideals is surjective.}
\item
A number field is {\em totally real} if every embedding is real,
i.e., $s=0$, and a number field is {\em totally complex} if every
embedding is complex, i.e., $r=0$.
\begin{enumerate}
\item (15 points) Find with proof the possible degrees of totally real fields.
\soln{
Every degree appears. The field $\Q$ is totally real, so it suffices
to prove that every degree $>1$ occurs.
Let $K=\Q(\zeta_n+1/\zeta_n)$ where~$n$ is a primitive
$n$th root of unity. Then for $3\geq 2$,~$K$ has degree $\vphi(n)/2$,
and is a Galois extension of~$\Q$ with cyclic Galois group, so there
is a subfield of~$K$ of degree~$d$ for each divisor~$d$ of $\vphi(n)/2$.
If $p^r$ is a prime power, then $p^r\mid \vphi(p^{r+1})$, etc., so we
see pretty easily that if $m$ is any integer then there exists $n$ such that
$m\mid \vphi(n)/2$.}
%(Your answer should be to state a subset of the positive integers and a proof
%that your answer is correct.)
\item (15 points) Find with proof the possible degrees of totally complex fields.
\soln{
Since $n=r+2s$, if $r=0$ then $n$ is even. In fact every even degree appears.
If $K$ is totally real and $i=\sqrt{-1}$, then $K(i)$ is totally complex,
since there is no real number $\alpha$ with $\alpha^2=-1$, so there's
no way to embed $K(i)$ into $\R$. The degree of $K(i)$ is twice the degree
of~$K$. As we saw in the previous part of this problem, every possible
degree occurs for totally real fields, so twice every degree occurs for
totally complex fields.
}
\end{enumerate}
\end{enumerate}
\end{document}
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